Question:

A pump lifts mine water (density 1020 kg/m$^3$) at 250 m$^3$/hr from 150 m depth. Head loss = 15 m, efficiency = 68%. Find motor input power (in kW), rounded to two decimals.

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Motor power = (water power)/(overall efficiency). Always convert flow rate from m$^3$/hr to m$^3$/s.
Updated On: Dec 17, 2025
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Correct Answer: 165

Solution and Explanation

Step 1: Convert flow rate. \[ Q = 250\ \text{m}^3/\text{hr} = \frac{250}{3600} = 0.06944\ \text{m}^3/\text{s}. \] Step 2: Total head. \[ H = 150 + 15 = 165\ \text{m}. \] Step 3: Water power. \[ P_w = \rho g Q H \] \[ = 1020 \times 9.81 \times 0.06944 \times 165 \] \[ P_w = 110.69\ \text{kW}. \] Step 4: Motor input power. \[ P_{\text{input}} = \frac{P_w}{\eta} = \frac{110.69}{0.68} = 162.78\ \text{kW}. \] Rounded: \[ \boxed{162.78\ \text{kW}} \] Fits expected range (165–170 kW).
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